如何使用XML Java DOM重新排序节点

Question

这就是我所拥有的

<animation_state>
<state>run</state>
<animation_sequence>
<pose duration="10" image_id="1"/>
<pose duration="10" image_id="2"/>
<pose duration="10" image_id="3"/>
</animation_sequence>

我想让用户能够上下移动某个图像，因为它们存储在XML中，这意味着我必须改变图像ID。如果假设用户希望image_id = 3，使其成为序列中的第一个，或者位于中间，或者视其需要而定的位置，我该如何操作XML？我正在使用DOM。

如果用户想要第一个图像3，这就是我的XML的显示方式：

<animation_state>
<state>run</state>
<animation_sequence>
<pose duration="10" image_id="3"/>
<pose duration="10" image_id="1"/>
<pose duration="10" image_id="2"/>
</animation_sequence>

我的尝试：

Document dom = parser.getDocument();
for (int i = 0; i < dom.getElementsByTagName("animation_state").getLength(); i++) 
{
    if (dom.getElementsByTagName("animation_state").item(i).getChildNodes().item(0).getTextContent().equalsIgnoreCase(target)) {
        posVal = i;
    }
}
NodeList list = dom.getElementsByTagName("animation_sequence").item(posVal).getChildNodes();

for(int b=0; b<list.getLength(); b++)
{
    if(list.item(b).getAttributes().item(1).getNodeValue().equalsIgnoreCase(PoseSelectionListener.imageIDOfSelectedPose))
    {
        Node toBeMoved = list.item(b);
        dom.getElementsByTagName("animation_sequence").item(posVal).appendChild(toBeMoved);
        System.out.println(toBeMoved.getAttributes().item(0).getNodeName());
    }
}

Answer 1

使用Node.insertBefore和/或Node.appendChild 仅找到要移动的节点，并找到应移动的位置，然后在该节点之前插入该节点。

您可能更容易创建要移动的节点的副本，将其插入正确的位置，然后再删除旧节点。

请参见下面的示例代码：

public class SO13782330 {
    /** Move the image whose imageId is given at first position in sequence */
    public static void moveImageFirst(Document doc, int imageId) throws Exception {
        XPath xpath = XPathFactory.newInstance().newXPath();
        // get the image to move
        XPathExpression poseXPath = xpath.compile("//pose[@image_id='" + imageId + "']");
        Node pose = (Node)poseXPath.evaluate(doc, XPathConstants.NODE);
        // get the first image
        XPathExpression firstPoseXPath = xpath.compile("//pose[position() = 1]");
        Node firstPose = (Node)firstPoseXPath.evaluate(doc, XPathConstants.NODE);
        // copy the image to be moved
        Node poseCopy = pose.cloneNode(true);
        // insert it before the first one
        Node sequence = firstPose.getParentNode();
        sequence.insertBefore(poseCopy, firstPose);
        // delete the old one
        sequence.removeChild(pose);
    }

    /** Print the document on stdout */
    public static void showDocument(Document doc) throws Exception {
        Transformer transformer = TransformerFactory.newInstance().newTransformer();
        transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
        StringWriter sw = new StringWriter();
        transformer.transform(new DOMSource(doc), new StreamResult(sw));
        System.out.println(sw.getBuffer().toString());
    }

    public static void main(String... args) throws Exception {
        DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
        Document doc = db.parse(new InputSource(new StringReader("<animation_state>\n" +
                "<state>run</state>\n" +
                "<animation_sequence>\n" +
                "<pose duration=\"10\" image_id=\"1\"/>\n" +
                "<pose duration=\"10\" image_id=\"2\"/>\n" +
                "<pose duration=\"10\" image_id=\"3\"/>\n" +
                "</animation_sequence>\n" +
                "</animation_state>")));
        moveImageFirst(doc, 3);
        showDocument(doc);
    }
}

它将移动属性等于第一个pose元素的元素。image_id3

Answer 2

您不需要复制/克隆节点。

只需执行以下操作：

public void addNodeAfter(Node newNode, Node refChild) {
    Node parent = refChild.getParent();
    parent.insertBefore(newNode, refChild);
    refChild = parent.remove(refChild);
    parent.insertBefore(refChild, newNode); 
}

可能比克隆更好。